How do you find all the real and complex roots of #x^7 + 128 = 0#?

2 Answers
Jan 12, 2016

Answer:

#x^7+128 = prod_(k = 0)^6 (x + 2cos((2kpi)/7) + 2isin((2kpi)/7))#

Hence roots:

#x = -2cos((2kpi)/7)-2isin((2kpi)/7)#

where #k=0,1,2,3,4,5,6#

Explanation:

#x^7+128 = x^7+2^7#

By De Moivre's formula we find:

#(cos((2kpi)/7) + i sin((2kpi)/7))^7 = cos(2kpi) + i sin(2kpi) = 1#

for any integer #k#.

So the #7#th roots of #1# can be written as:

#cos((2kpi)/7) + i sin((2kpi)/7)# where #k=0,1,2,3,4,5,6#

Hence we find:

#x^7+2^7 = prod_(k = 0)^6 (x + 2cos((2kpi)/7) + 2isin((2kpi)/7))#

Jan 13, 2016

Answer:

#z = 2cis(pi + (2pik)/7)# for #k = 0,1,2,3,4,5,6#

Explanation:

We have

#x^7 + 128 = 0#

Numerically we know that #-2# is one of the roots.
#z = -2# has modulo #2# and #theta = pi#, so we have the trigonometric representation

#z = 2cis(pi)#

(Where #cis(theta) = cos(theta) + isin(theta)#)

But the thing is, whenever we take the nth root of a number, we bissect the circle of #root(n)(|z|)# radius in #n# parts of equal distance. I.e.: if we take the square roots, the two roots will be exactly #(2pi)/2# degrees apart, if we take the cubic root, they'll be #(2pi)/3# apart and so on.

That means that we can derive that the general form of the roots are

#z = 2cis(pi + (2pik)/7)# for #k = 0,1,2,3,4,5,6# as when we have #k = 7#, we go back to the same root we already had.

Which we can easily check back using De Moivre's formula, since

#z^7 = 128cis(7pi + 7(2pik)/7) = 128cis(7pi + 2pik)#

For any of the mentioned #k#, adding or subtracting #2pi# is the same as doing nothing so

#z^7 = 128cis(7pi)#

And since #7pi = 6pi + pi#

#z^7 = 128cis(pi) = -128#